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'I.isted in the same order as the reactants in Table 3.2. ' A negative sign implies that the constituent is receiving electrons. 'By definition, oxygen demand is negative oxygen.

'I.isted in the same order as the reactants in Table 3.2. ' A negative sign implies that the constituent is receiving electrons. 'By definition, oxygen demand is negative oxygen.

Example 3.1.1.1

Consider a typical molar-based stoichiometric equation for bacterial growth on carbohydrate (CHO) with ammonia as the nitrogen source:

CH O ■+ 0.290 O. + 0.142 NH4 + 0.142 HCO, ->■

where C,H O N is the empirical formula for cell mass. Note that the charges are balanced and that the number of moles of each element in the reactants equals the number in the products. The molar-based stoichiometric equation tells us that the biomass yield is 0.142 moles of biomass formed per mole of carbohydrate used and that 0.290 moles of oxygen are required per mole of carbohydrate used to synthesize that biomass.

Convert this equation to a mass based stoichiometric equation. To do this, we need the molecular weight of each reactant and product. These are CH-O. 30; O , 32; NH, . 18; HCO,. 61; C,H-O..N, I 13; CO.., 44; and H O, 18. Using these with the stoichiometric coefficients from Eq. 3.6 in F.q. 3.2 gives:

In this case, the charges are no longer balanced, but the sum of the stoichiometric coefficients for the reactants equals the sum for the products. The mass-based stoichiometric equation tells us that the biomass yield is 0.535 grams of biomass formed per gram of carbohydrate used and that 0.309 grams of oxygen are required per gram of carbohydrate used to synthesize that biomass.

Now convert the molar-based equation to a COD-based equation. To do this, use must be made of the unit CODs given in Table 3.1. In this case, the unit COD of ammonia is taken as zero because the nitrogen in cell material is in (he same oxidation state as the nitrogen in ammonia, i.e., -Ill: it does not undergo a change of oxidation state. Carrying out the conversion represented by Eq. 3.4 yields:

Note that only three constituents remain because they are the only ones that can be represented by COD in this case. Also note that like the mass-based equation, the sum of the stoichiometric coefficients for the reactants equals the sum of the stoichiometric coefficients for the products. Finally, note that the stoichiometric coefficient for oxygen carries a negative sign even though it is a reactant. That is because it is being expressed as COD. Thus, the COD-based stoichiometric equation tells us that the biomass yield is 0.71 grams of biomass COD formed per gram of carbohydrate COD used and that 0.29 grams of oxygen are required per gram of carbohydrate COD used to synthesize that biomass.

3.1.2 Generalized Reaction Rate

Stoichiometric equations can also be used to establish the relative reaction rates for reactants or products. Because the sum of the stoichiometric coefficients in a mass-based stoichiometric equation equals zero, its general form may be rewritten in the following way:""

where components 1 through k are reactants, components k+1 through m are products, and reactant A, is the basis for the normalized stoichiometric coefficients. Note that the normalized stoichiometric coefficients are given negative signs for reactants and positive signs for products. Since there is a relationship between the masses of the different reactants used or products formed, it follows that there is also a relationship between the rates at which they are used or formed. If we let r, represent the rate of formation of component i (where i = 1 —» k), it follows that:

where r is called the generalized reaction rate. As above, the sign on ty, signifies whether the component is being removed or formed. Consequently, if the stoichiometry of a reaction has been determined in mass units, and the reaction rate has been determined for one component, then the reaction rates in mass units are known for all other components.

Example 3.1.2.1

Biomass is growing in a bioreactor at a rate of 1.0 g/(L h) and the growth conforms to the stoichiometry expressed by Eq. 3.7. At what rate are carbohydrate and oxygen being used in the bioreactor to support that growth? Rewriting Eq. 3.7 in the form of Eq. 3.9 gives:

Use of Eq. 3.10 allows determination of the generalized reaction rate: